To solve the problem step by step, we will first find the coordinates of the vertices of triangle ABC using the given midpoints D, E, and F. Then, we will calculate the centroid G of triangle ABC and finally compute the required expression.
### Step 1: Use the midpoint formula to find the coordinates of A, B, and C.
Given:
- D(-2, 3) is the midpoint of BC
- E(4, -3) is the midpoint of CA
- F(4, 5) is the midpoint of AB
Using the midpoint formula:
- For midpoint D:
\[
D = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) = (-2, 3)
\]
This gives us two equations:
\[
\frac{x_2 + x_3}{2} = -2 \implies x_2 + x_3 = -4 \quad (1)
\]
\[
\frac{y_2 + y_3}{2} = 3 \implies y_2 + y_3 = 6 \quad (2)
\]
- For midpoint E:
\[
E = \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right) = (4, -3)
\]
This gives us:
\[
\frac{x_1 + x_3}{2} = 4 \implies x_1 + x_3 = 8 \quad (3)
\]
\[
\frac{y_1 + y_3}{2} = -3 \implies y_1 + y_3 = -6 \quad (4)
\]
- For midpoint F:
\[
F = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (4, 5)
\]
This gives us:
\[
\frac{x_1 + x_2}{2} = 4 \implies x_1 + x_2 = 8 \quad (5)
\]
\[
\frac{y_1 + y_2}{2} = 5 \implies y_1 + y_2 = 10 \quad (6)
\]
### Step 2: Solve the system of equations.
Now we have the following equations:
1. \( x_2 + x_3 = -4 \)
2. \( y_2 + y_3 = 6 \)
3. \( x_1 + x_3 = 8 \)
4. \( y_1 + y_3 = -6 \)
5. \( x_1 + x_2 = 8 \)
6. \( y_1 + y_2 = 10 \)
From equations (1) and (3):
- From (1): \( x_3 = -4 - x_2 \)
- Substitute into (3):
\[
x_1 + (-4 - x_2) = 8 \implies x_1 - x_2 = 12 \quad (7)
\]
From equations (5) and (7):
- Adding (5) and (7):
\[
(x_1 + x_2) + (x_1 - x_2) = 8 + 12 \implies 2x_1 = 20 \implies x_1 = 10
\]
- Substitute \( x_1 \) back into (5):
\[
10 + x_2 = 8 \implies x_2 = -2
\]
- Substitute \( x_2 \) into (1):
\[
-2 + x_3 = -4 \implies x_3 = -2
\]
Now we have \( x_1 = 10, x_2 = -2, x_3 = -2 \).
For the y-coordinates:
- From (2) and (4):
\[
y_3 = 6 - y_2 \quad (8)
\]
- Substitute into (4):
\[
y_1 + (6 - y_2) = -6 \implies y_1 - y_2 = -12 \quad (9)
\]
- From (6):
\[
y_1 + y_2 = 10 \quad (10)
\]
Adding (9) and (10):
\[
(y_1 - y_2) + (y_1 + y_2) = -12 + 10 \implies 2y_1 = -2 \implies y_1 = -1
\]
Substituting back:
\[
-1 + y_2 = 10 \implies y_2 = 11
\]
Substituting into (8):
\[
y_3 = 6 - 11 = -5
\]
Now we have the coordinates of A, B, and C:
- A(10, -1)
- B(-2, 11)
- C(-2, -5)
### Step 3: Find the centroid G of triangle ABC.
The centroid G is given by:
\[
G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)
\]
Substituting the coordinates:
\[
G = \left( \frac{10 + (-2) + (-2)}{3}, \frac{-1 + 11 + (-5)}{3} \right) = \left( \frac{6}{3}, \frac{5}{3} \right) = (2, \frac{5}{3})
\]
### Step 4: Calculate the distances AG, BG, and CG.
Using the distance formula:
\[
AG = \sqrt{(x_1 - x_G)^2 + (y_1 - y_G)^2} = \sqrt{(10 - 2)^2 + (-1 - \frac{5}{3})^2}
\]
Calculating:
\[
AG = \sqrt{8^2 + \left(-1 - \frac{5}{3}\right)^2} = \sqrt{64 + \left(-\frac{8}{3}\right)^2} = \sqrt{64 + \frac{64}{9}} = \sqrt{\frac{576 + 64}{9}} = \sqrt{\frac{640}{9}} = \frac{8\sqrt{10}}{3}
\]
Similarly, calculate BG and CG:
\[
BG = \sqrt{(-2 - 2)^2 + (11 - \frac{5}{3})^2} = \sqrt{(-4)^2 + \left(\frac{28}{3}\right)^2} = \sqrt{16 + \frac{784}{9}} = \sqrt{\frac{144 + 784}{9}} = \sqrt{\frac{928}{9}} = \frac{4\sqrt{58}}{3}
\]
\[
CG = \sqrt{(-2 - 2)^2 + (-5 - \frac{5}{3})^2} = \sqrt{(-4)^2 + \left(-\frac{20}{3}\right)^2} = \sqrt{16 + \frac{400}{9}} = \sqrt{\frac{144 + 400}{9}} = \sqrt{\frac{544}{9}} = \frac{4\sqrt{34}}{3}
\]
### Step 5: Calculate the final expression.
Now we need to find:
\[
\sqrt{|AG|^2 + |BG|^2 - |CG|^2}
\]
Calculating:
\[
|AG|^2 = \left(\frac{8\sqrt{10}}{3}\right)^2 = \frac{640}{9}
\]
\[
|BG|^2 = \left(\frac{4\sqrt{58}}{3}\right)^2 = \frac{928}{9}
\]
\[
|CG|^2 = \left(\frac{4\sqrt{34}}{3}\right)^2 = \frac{544}{9}
\]
Now substituting:
\[
\sqrt{\frac{640}{9} + \frac{928}{9} - \frac{544}{9}} = \sqrt{\frac{640 + 928 - 544}{9}} = \sqrt{\frac{1024}{9}} = \frac{32}{3}
\]
### Final Answer:
\[
\frac{32}{3}
\]
